# EM waves: seeming contradiction in vacuum assumption

It can be shown starting from Maxwell's equations that the electromagnetic field satisfies the wave equation: $$\square^2 \mathbf{E} = \frac{1}{\epsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf{J}}{\partial t}$$ $$\square^2 \mathbf{B}=\mu_0\nabla\times\mathbf{J}$$ In particular, when the RHS of the above equations are zero we may write the fields as plane waves.

As a specific example, consider Example 2 on page 402 in Griffiths: $\tilde{\mathbf{E}}=\tilde{E}_0e^{i(kz-\omega t)}\hat{\mathbf{x}}$ , $\tilde{\mathbf{B}}=\frac{1}{c}\tilde{E}_0e^{i(kz-\omega t)}\hat{\mathbf{y}}$. Taking the divergence of the first and curl of the second then gives $$\nabla\cdot\tilde{\mathbf{E}}=0$$ $$\nabla\times\tilde{\mathbf{B}}=-i\frac{k}{c}\tilde{E}_0e^{i(kz-\omega t)}\hat{\mathbf{x}}$$ From which we may identify the RHS of the second to be $\mu_0\mathbf{J}$. This is clearly not a vacuum in the strict sense, which is one seeming contradiction but the chapter is not focused on the space in which waves live, so it is more or less negligible.

Inspection of the wave equations show that homogeneity is still satisfied if $\rho =\epsilon_0\partial_t\lambda$ and $\mathbf{J}=\nabla\lambda$ for some $\lambda$. The second contradiction manifests in calculating $\lambda = -i\frac{k}{c}\tilde{E}_0e^{i(kz-\omega t)}x + f(t,y,z)$; there is no $\rho$, and $f$ cannot depend on $x$, so the first wave equation is not homogenous.

Is the above analysis correct, and if so can these contradictions be reconciled? In other words, can some solution both agree with the vacuum Maxwell equations (or at least homogenous wave equations) and preserve the general analysis of electromagnetic waves?

• Where do your first two equations come from? Jul 1 '18 at 4:16
• @RobJeffries, The first two equations are a result from the maxwell equations (not assuming vacuum) by taking the curl of curl then simplifying.
– user195162
Jul 1 '18 at 4:46

What you have done wrong is identify: $$\nabla \times \vec B$$ with $\mu_0 \vec J$. You are missing Maxwell's addition: $$\nabla \times \vec B=\mu_0 \vec J+c\frac{\partial \vec E}{\partial t}$$ subbing the $\vec B$ and $\vec E$ field you have written into these will give: $$\vec J=0$$